Đáp án:

\(\begin{array}{l}
a){\left( {a - 1} \right)^2}\\
\end{array}\) 

b) \(\left[ \begin{array}{l}
1 \le a < 3 - \sqrt 3 \\
1 > a >  - 1 + \sqrt 3 
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:a \ge 0;a \ne 1\\
P = \left[ {\dfrac{{\left( {1 - \sqrt a } \right)\left( {a + \sqrt a  + 1} \right)}}{{1 - \sqrt a }} + \sqrt a } \right].\left[ {\dfrac{{\left( {1 + \sqrt a } \right)\left( {a - \sqrt a  + 1} \right)}}{{1 + \sqrt a }} - \sqrt a } \right]\\
 = \left( {a + \sqrt a  + 1 + \sqrt a } \right).\left( {a - \sqrt a  + 1 - \sqrt a } \right)\\
 = {\left( {\sqrt a  + 1} \right)^2}.{\left( {\sqrt a  - 1} \right)^2}\\
 = {\left( {a - 1} \right)^2}\\
b)P < 7 - 4\sqrt 3 \\
 \to {\left( {a - 1} \right)^2} < 7 - 4\sqrt 3 \\
 \to {\left( {a - 1} \right)^2} < 4 - 2.2.\sqrt 3  + 3\\
 \to {\left( {a - 1} \right)^2} < {\left( {2 - \sqrt 3 } \right)^2}\\
 \to \left| {a - 1} \right| < \left| {2 - \sqrt 3 } \right|\\
 \to \left[ \begin{array}{l}
a - 1 < 2 - \sqrt 3 \left( {DK:a \ge 1} \right)\\
a - 1 >  - 2 + \sqrt 3 \left( {DK:a < 1} \right)
\end{array} \right.\\
\end{array}\)

\( \to \left[ \begin{array}{l}
1 \le a < 3 - \sqrt 3 \\
1 > a >  - 1 + \sqrt 3 
\end{array} \right.\)

Đáp án:

`a.`Rút gọn:

`P=(\frac{1-asqrta}{1-sqrta}+sqrta)(\frac{1+asqrta}{1+sqrta}-sqrta)`

`= (\frac{(1-sqrta)(1+sqrta+a)}{1-sqrta}+sqrta)(\frac{(1+sqrta)(1-sqrta+a)}{1+sqrta}-sqrta)`

`= (1+sqrta+a+sqrta)(1-sqrta+a-sqrta)`

`= (1+2sqrta+a)(1-2sqrta+a)`

`= (1+a)^2-4a`

`= 1+2a+a^2-4a`

`= 1-2a+a^2`

`= a^2-2a+1`

`b.`Tìm `a`:

Để `P<7-4sqrt3` thì `a^2-2a+1<7-4sqrt3`

`<=> (a-1)^2<7-4sqrt3`

`<=> |a-1|<sqrt{7-4sqrt3}`

`<=> |a-1|<sqrt{(2-sqrt3)^2}`

`<=> |a-1|<2-sqrt3`

`<=>`\(\left[ \begin{array}{l}a-1<2\sqrt{3}\ (a\ge1)\\-(a-1)<2-\sqrt{3}\ (a<1)\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}a<3-\sqrt{3} \ (a\ge1)\\a>-1+\sqrt{3} \ (a<1)\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}1\le a<3-\sqrt{3}\\-1+\sqrt{3}<x<1\end{array} \right.\) 

`<=> -1+sqrt3<a<3-sqrt3`